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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p>From (<a href="" class="xref" data-knowl="./knowl/eq5_6.html" title="Equation 5.3.3">(5.3.3)</a>):</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_6.html ./knowl/eq5_7.html">
\begin{equation*}
y_1(x)=\sum_{n=0}^{\infty} a_n x^n, \quad \textrm{with}~ a_0=1, a_1=0.
\end{equation*}
</div>
<p class="continuation">In (<a href="" class="xref" data-knowl="./knowl/eq5_7.html" title="Equation 5.3.4">(5.3.4)</a>), let <span class="process-math">\(a_0=1, a_1=0\text{.}\)</span> Then</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_6.html ./knowl/eq5_7.html">
\begin{equation*}
\begin{aligned}
n=0: &amp;\quad a_2=-\frac{\alpha (\alpha+1)}{2!}, \quad (k=1)\\
n=1: &amp; \quad a_3=0,\\
n=2: &amp;\quad a_4=-\frac{(\alpha-2) (\alpha+3)}{4 \cdot 3} a_2=\frac{\alpha (\alpha-2) (\alpha+1) (\alpha+3)}{4!}, \quad (k=2)\\
n=3: &amp;\quad a_5=0,\\
n=4: &amp;\quad a_6=-\frac{(\alpha-4) (\alpha+5)}{6 \cdot 5} a_4=-\frac{\alpha (\alpha-2) (\alpha-4) (\alpha+1) (\alpha+3) (\alpha+5)}{6!},\quad (k=3)\\
&amp;\quad \cdots\\
&amp;\quad a_{2 k+1}=0,\\
&amp; \quad a_{2 k}=(-1)^k \frac{\alpha (\alpha-2) (\alpha-4) \cdots (\alpha-2 (k-1)) (\alpha+1) (\alpha+3) (\alpha+5) \cdots (\alpha+2 k-1)}{(2 k)!}, \quad k \geq 1.
\end{aligned}
\end{equation*}
</div>
<p class="continuation">Thus,</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_6.html ./knowl/eq5_7.html">
\begin{equation*}
\begin{aligned}
y_1=\sum_{n=0}^{\infty} a_n x^n&amp;=1+\sum_{k=1}^{\infty} a_{2 k} x^{2 k}+\sum_{k=0}^{\infty} a_{2k+1} x^{2k+1}\\
&amp;=1+\sum_{k=1}^{\infty} (-1)^k \frac{\alpha (\alpha-2) \cdots (\alpha-2k+2) (\alpha+1)(\alpha+3)\cdots (\alpha+2k-1)}{(2k)!} x^{2k}.
\end{aligned}
\end{equation*}
</div>
<span class="incontext"><a href="sec5_3.html#p-222" class="internal">in-context</a></span>
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